Step 1

Since you have asked multiple question, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.

Step 2

We Know that Synthetic Division is a method to divide a polynomial by a linear expression.

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}+{\left(-{3}-{2}{i}\right)}{x}^{{{2}}}+{\left(-{3}-{i}\right)}{x}+{\left({2}+{i}\right)}\)

by synthetic division.

we have given 2+i is factor (zero) of function

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{2}+{i}&{2}&-{3}-{2}{i}&-{3}-{i}&{2}+{i}\backslash&&{4}+{2}{i}&{2}+{i}&-{2}-{i}\backslash{h}{l}\in{e}-{1}&{2}&{1}&-{1}&{0}\backslash&&-{2}&{1}\backslash{h}{l}\in{e}&{2}&-{1}&{0}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

So, f(x) can be written in its linear form at.

\(\displaystyle{\left({x}-{\left({2}+{i}\right)}{\left({x}+{1}\right)}{\left({2}{x}-{1}\right)}={0}\right.}\)

[Here \(\displaystyle{2}+{i},-{1},{\frac{{{1}}}{{{2}}}}\) are zero of function]

Since you have asked multiple question, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.

Step 2

We Know that Synthetic Division is a method to divide a polynomial by a linear expression.

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}+{\left(-{3}-{2}{i}\right)}{x}^{{{2}}}+{\left(-{3}-{i}\right)}{x}+{\left({2}+{i}\right)}\)

by synthetic division.

we have given 2+i is factor (zero) of function

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{2}+{i}&{2}&-{3}-{2}{i}&-{3}-{i}&{2}+{i}\backslash&&{4}+{2}{i}&{2}+{i}&-{2}-{i}\backslash{h}{l}\in{e}-{1}&{2}&{1}&-{1}&{0}\backslash&&-{2}&{1}\backslash{h}{l}\in{e}&{2}&-{1}&{0}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

So, f(x) can be written in its linear form at.

\(\displaystyle{\left({x}-{\left({2}+{i}\right)}{\left({x}+{1}\right)}{\left({2}{x}-{1}\right)}={0}\right.}\)

[Here \(\displaystyle{2}+{i},-{1},{\frac{{{1}}}{{{2}}}}\) are zero of function]